Electric Field Help: Questions & Answers

In summary: The electric field at a distance of 8.8 cm from the cylindar’s center is then:The electric field at a distance of 8.8 cm from the cylindar’s center is then:In summary, the electric field at a distance of 8.8 cm from the cylindar’s center is 1.68 × 10^2 N/C.
  • #1
arutha
6
0
I have these questions on electric fields that I'm a bit confused on..

A flat circle of radius 8 cm is placed in a uniform electric field of 8.5 × 10^2 N/C. What is the electric flux (in Nm^2/C) through the circle when its face is at 51° to the field lines?

I just use EAcos(theta) don't I? Where A is 2*pi*r, but that angle do I use 51 or 90-51 since it is the angle is meant to be between the normal and the object not the object and the surface right?

A metallic sphere of radius 22 cm is negatively charged. The magnitude of the resulting electric field, close to the outside surface of the sphere, is 1.8 × 10^2 N/C. Calculate the net electric flux (in Nm^2/C) outward through a spherical surface surrounding, and just beyond, the metallic sphere's surface.

I'm thinking just E*A*cos(theta) again.. Would the answer be negative because it is negatively charged?


Two concentric spherical shells of radii R1=1 m and R2=2 m, contain charge Q1=0.005 C and Q2=0.0065 C respectively.
Calculate the Electric field at a distance r=1.79 m from the centerpoint of the spheres


I have absolutely no idea on this one.. How does it work with the two charges? And what if I was calculating the field outside the two spheres, would that be any different?

A very long solid nonconducting cylinder of radius 18.3 cm possesses a uniform volume charge density of 1.68 μC/m^3. Determine the magnitude of the electric field (in N/C) inside the cylinder at a radial distance of 8.8 cm from the cylinder's central axis

Heres what I've thought of, multiply the volume charge density by the volume of the cylindar to get the charge in μC, then use E=kQ/r^2 to get the magnitude of the electric field. Is that right? Edit: That won't work because I don't have a length of the cylindar to get the volume... Woops.

Thanks for any help, btw I don't want numbers or any answers I'd rather hear the process then get the numbers myself so I can figure out other problems of similar nature..
 
Last edited:
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  • #2
oooooooooooo
what is the area of a circle 2*pi*r ? Check it.
 
  • #3
Oh yeah I forgot the square after the r... I wrote it down on the sheet, just missed typing it.
 
  • #4
A = (Pi)*r^2
 
  • #5
For the rest problems go through Guass's Theorem
 
  • #6
Well, I got them all except the last one now. Still have absolutely no idea how to do it, I've gone through my textbook, lecture notes and everythnig but can't find anything on it.
 
  • #7
arutha said:
A very long solid nonconducting cylinder of radius 18.3 cm possesses a uniform volume charge density of 1.68 μC/m^3. Determine the magnitude of the electric field (in N/C) inside the cylinder at a radial distance of 8.8 cm from the cylinder's central axis.

The distance of the point at which the field magnitude is required is 8.8 cm which is less then the radius of cylinder 18.3 cm.
Consider a coaxial cylindrical Gaussian surface of radius 8.8 cm and apply the Gauss’s theorem. Remember the charge to be taken within the Gaussian surface.
 

What is an electric field?

An electric field is a region in space where electrically charged particles experience a force. It is created by electric charges and can be either positive or negative.

What is the unit of electric field?

The unit of electric field is newtons per coulomb (N/C) in the SI system of units. In the CGS system, it is expressed as dynes per statcoulomb (dyn/cm^2).

How is electric field strength calculated?

The electric field strength is calculated by dividing the force experienced by an electric charge by the magnitude of the charge. In mathematical terms, it is represented as E = F/q, where E is the electric field strength, F is the force, and q is the charge.

What is the difference between electric field and electric potential?

Electric field is a vector quantity that represents the force experienced by a charged particle, while electric potential is a scalar quantity that represents the energy per unit charge at a specific point in an electric field.

How is electric field related to electric potential?

Electric field and electric potential are related by the equation E = -dV/dr, where E is the electric field, V is the electric potential, and r is the distance from the source of the electric field. In other words, the electric field is the negative gradient of the electric potential.

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